Answer
$\varnothing$.
Work Step by Step
In the given equation we can write the divisor as
$\Rightarrow \left ( \frac{1}{x+1}+\frac{x}{1-x} \right ) \div \left ( \frac{x}{x+1}-\frac{1}{-1(1-x)} \right ) =-1$
$\Rightarrow \left ( \frac{1}{x+1}+\frac{x}{1-x} \right ) \div \left ( \frac{x}{x+1}+\frac{1}{1-x} \right ) =-1$
The LCD is $(x+1)(1-x)$.
Multiply the numerator and the denominator to form LCD at the denominators.
$\Rightarrow \left ( \frac{1-x}{(x+1)(1-x)}+\frac{x(x+1)}{(x+1)(1-x)} \right ) \div \left ( \frac{x(1-x)}{(x+1)(1-x)}+\frac{x+1}{(x+1)(1-x)} \right ) =-1$
Add numerators because denominators are equal.
$\Rightarrow \frac{1-x+x(x+1)}{(x+1)(1-x)} \div \frac{x(1-x)+x+1}{(x+1)(1-x)} =-1$
Invert the divisor and multiply.
$\Rightarrow \frac{1-x+x(x+1)}{(x+1)(1-x)} \cdot \frac{(x+1)(1-x)}{x(1-x)+x+1} =-1$
Cancel common factors.
$\Rightarrow \frac{1-x+x(x+1)}{x(1-x)+x+1} =-1$
Use the distributive property.
$\Rightarrow \frac{1-x+x^2+x}{x-x^2+x+1} =-1$
Add like terms.
$\Rightarrow \frac{1+x^2}{-x^2+2x+1} =-1$
Multiply the equation by $(-x^2+2x+1)$.
$\Rightarrow (-x^2+2x+1)\cdot \frac{1+x^2}{-x^2+2x+1} =-1(-x^2+2x+1)$
Simplify.
$\Rightarrow 1+x^2=x^2-2x-1$
Add $-x^2+2x-1$ to both sides.
$\Rightarrow 1+x^2-x^2+2x-1=x^2-2x-1-x^2+2x-1$
Simplify.
$\Rightarrow 2x=-2$
Divide both sides by $2$.
$\Rightarrow \frac{2x}{2}=\frac{-2}{2}$
Simplify.
$\Rightarrow x=-1$
The original equation is undefined for $x=-1$ because $-1$ is a zero of a denominator.\.
There is no solution or $\varnothing$.
